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cái này bảo tìm GT \(\Rightarrow\) P có GT cố định
ta có : \(a=b=c=1\) thỏa mãn đk bài toán
thế vào P ta có \(P=0\)
Bạn ghi đề sai rồi hèn chi giải chả ra!
\(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Ta có:
\(a^3+b^3+c^3=3abc=>a^3+b^3+c^3-3abc=0\)
\(=>\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3a^2b-3ab^2-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Vì a3+b3+c3=3abc và a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
\(=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Tổng 3 số không âm = 0 <=> chúng đều = 0
\(< =>\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}< =>a=b=c}\)
Vậy \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
\(\)
Ta có ; \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\frac{a+b+c}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vì \(a+b+c\ne0\) nên ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
a) Thay a = b = c vào biểu thức được : \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
b) Thay a = b = c vào P : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)
Ta có: \(a^3+b^3+c^3-3abc=0\) \(\Leftrightarrow a+b+c=0\) hoặc a = b = c
theo gt thi a + b + c \(\ne0\) \(\Rightarrow a=b=c\)
\(\Rightarrow N=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Hoặc \(a+b+c=0\)
Hoặc \(\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
TH1 : \(a+b+c=0\Rightarrow a=-\left(b+c\right);b=-\left(a+c\right);c=-\left(a+b\right)\)
\(\Rightarrow\)\(A=\left[1-\frac{\left(b+c\right)}{b}\right]\left[1-\frac{\left(a+c\right)}{c}\right]\left[1-\frac{\left(a+b\right)}{a}\right]\)
\(\Rightarrow\)\(A=\left(1-1-\frac{c}{b}\right)\left(1-1-\frac{a}{c}\right)\left(1-1-\frac{b}{a}\right)\)
\(\Rightarrow\)\(A=\left(\frac{-c}{b}\right)\left(\frac{-a}{c}\right)\left(\frac{-b}{a}\right)=-1\)
TH2 : \(\left(a^2+b^2+c^2-ab-bc-ac\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\)\(a-b=b-c=c-a=0\)hay \(a=b=c=0\)
\(\Rightarrow\)\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
1.CMR:
a) 3.\(\left(x^2+y^2+z^2\right)-\left(x-y\right)^2\) \(-\left(y-z\right)^2-\left(z-x\right)^2=\left(x+y+z\right)^2\)
Ta có : \(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
\(\Leftrightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3=3ab.bc.ac\)
Đặt \(ab=x;bc=y;ac=z\) . Khi đó , ta có :
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x^3+y^3+3x^2y+3y^2x\right)+z^3-3x^2y-3y^2x-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right]=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+z=0\\x^2+y^2+z^2-xy-yz-xz=0\end{matrix}\right.\)
Với \(x+y+z=0\Rightarrow ab+ac+bc=0\)
Với \(x^2+y^2+z^2-xy-yz-xz=0\)
\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
Lí luận tổng này \(\ge0\) ( làm tắt )
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\y=z\end{matrix}\right.\) \(\Rightarrow x=y=z\)
\(\Rightarrow ab=ac=bc\)
....
Đến bước này chịu , bạn xem đề có sai không ?
ĐK: \(abc\ne0\)
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2+c^2-c\left(a+b\right)\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(A=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{a+c}{a}\right)=\frac{\left(-a\right).\left(-b\right)\left(-c\right)}{abc}=-1\)
TH2: \(a=b=c\)
\(\Rightarrow A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Giải:
Từ \(a^3+b^3+c^3=3abc\Leftrightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Ta xét các trường hợp:
Trường hợp \(1\): Nếu \(a+b+c=0\) thì:
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Thay vào \(P\) ta có:
\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{c}\right)\)
\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{\cdot\left(-c\right).\left(-a\right).\left(-b\right)}{b.c.a}=-1\)
Trường hợp \(2\): Nếu \(a=b=c\) thì:
\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
\(=2.2.2=8\)
Vậy \(P=-1\) hoặc \(P=8\)
ta có : a3+b3+c3-3abc=0
\(\Rightarrow\)(a+b)3+c3-3abc-3a2b-3ab2=0
\(\Rightarrow\)(a+b+c)(a2+b2+c2+2ab-ac-bc)-3ab(a+b+c)=0
\(\Rightarrow\)(a+b+c)(a2+b2+c2-ab-ac-bc)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\\\left(a+b+c\right)^2+a^2+b^2+c^2=0\Leftrightarrow a=b=c=0\left(bỏ\right)\end{matrix}\right.\)ta có P=(1+\(\dfrac{a}{b}\))(1+\(\dfrac{b}{c}\))(1+\(\dfrac{c}{a}\))
\(\Leftrightarrow\)p=\(\left(\dfrac{b+a}{b}\right)\left(\dfrac{c+b}{c}\right)\left(\dfrac{a+c}{a}\right)\)
\(\Leftrightarrow P=\left(\dfrac{-c}{b}\right)\left(\dfrac{-a}{c}\right)\left(\dfrac{-b}{a}\right)\)
\(\Leftrightarrow\)P=-1